Solve $x^3-y^3-z^3=3^t cdot 2xyz$
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Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$
We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.
I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.
number-theory diophantine-equations
add a comment |Â
up vote
6
down vote
favorite
Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$
We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.
I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.
number-theory diophantine-equations
This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
â Right
Aug 30 at 6:29
@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
â Oleg567
Aug 30 at 6:34
@Oleg567 That is right, did not see it immediately.
â Right
Aug 30 at 6:36
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$
We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.
I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.
number-theory diophantine-equations
Solve the diophantine equation $x^3-y^3-z^3=3^t cdot 2xyz,(x,y,z in Bbb N)$, where $tin Bbb N.$
We can find infinitely many solutions for $t=1$ from initial solution $x=52,y=21,z=19$.
I search for all $x<13000$ but find no solution for $t>1$, I want to know if there are any solutions for $t>1$.
number-theory diophantine-equations
number-theory diophantine-equations
edited Aug 30 at 5:55
asked Aug 30 at 2:45
Next
8,46932269
8,46932269
This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
â Right
Aug 30 at 6:29
@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
â Oleg567
Aug 30 at 6:34
@Oleg567 That is right, did not see it immediately.
â Right
Aug 30 at 6:36
add a comment |Â
This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
â Right
Aug 30 at 6:29
@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
â Oleg567
Aug 30 at 6:34
@Oleg567 That is right, did not see it immediately.
â Right
Aug 30 at 6:36
This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
â Right
Aug 30 at 6:29
This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
â Right
Aug 30 at 6:29
@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
â Oleg567
Aug 30 at 6:34
@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
â Oleg567
Aug 30 at 6:34
@Oleg567 That is right, did not see it immediately.
â Right
Aug 30 at 6:36
@Oleg567 That is right, did not see it immediately.
â Right
Aug 30 at 6:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.
Some solutions for small $t$'s.
beginarrayc
hline
t& u & v & w \ hline
2 & 35917476 & 6829645 & 10182731\ hline
3 & 1137565 & 647349 & 30196 \ hline
4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
endarray
It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.
for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.
For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)
In particular at $t=6$ the original curve
$$
frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
$$
is birational to
$$
y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
$$
via
$$
beginalign
(u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
endalign
$$
We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.
Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.
$t=2$ example
For $t=2$, we want integer solutions to
$$
f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
$$
We may freely multiply a rational $tneq 0$ to get
$$
tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
$$
Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).
Let
$$
beginalign
t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
endalign
$$
then it may be verified that this transform the original equation to
$$
g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
$$
which is a projective elliptic curve.
(In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)
Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
$$
E: y^2 = x^3-36288x+2285712
$$
This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.
Sample solutions.
The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
$$
(-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
$$
Converting to $(u,v,w)$, we get
$$
(frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
$$
Normalizing, the first three solutions are
$$
beginalign
(95,-42,-13),\
(-35917476, -6829645, -10182731)\
(54059017558123943, -5902844861231317, -35013190193908290)
endalign
$$
Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
$$
(35917476, 6829645, 10182731)
$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.
Some solutions for small $t$'s.
beginarrayc
hline
t& u & v & w \ hline
2 & 35917476 & 6829645 & 10182731\ hline
3 & 1137565 & 647349 & 30196 \ hline
4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
endarray
It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.
for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.
For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)
In particular at $t=6$ the original curve
$$
frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
$$
is birational to
$$
y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
$$
via
$$
beginalign
(u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
endalign
$$
We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.
Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.
$t=2$ example
For $t=2$, we want integer solutions to
$$
f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
$$
We may freely multiply a rational $tneq 0$ to get
$$
tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
$$
Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).
Let
$$
beginalign
t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
endalign
$$
then it may be verified that this transform the original equation to
$$
g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
$$
which is a projective elliptic curve.
(In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)
Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
$$
E: y^2 = x^3-36288x+2285712
$$
This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.
Sample solutions.
The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
$$
(-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
$$
Converting to $(u,v,w)$, we get
$$
(frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
$$
Normalizing, the first three solutions are
$$
beginalign
(95,-42,-13),\
(-35917476, -6829645, -10182731)\
(54059017558123943, -5902844861231317, -35013190193908290)
endalign
$$
Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
$$
(35917476, 6829645, 10182731)
$$
add a comment |Â
up vote
3
down vote
accepted
Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.
Some solutions for small $t$'s.
beginarrayc
hline
t& u & v & w \ hline
2 & 35917476 & 6829645 & 10182731\ hline
3 & 1137565 & 647349 & 30196 \ hline
4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
endarray
It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.
for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.
For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)
In particular at $t=6$ the original curve
$$
frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
$$
is birational to
$$
y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
$$
via
$$
beginalign
(u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
endalign
$$
We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.
Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.
$t=2$ example
For $t=2$, we want integer solutions to
$$
f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
$$
We may freely multiply a rational $tneq 0$ to get
$$
tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
$$
Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).
Let
$$
beginalign
t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
endalign
$$
then it may be verified that this transform the original equation to
$$
g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
$$
which is a projective elliptic curve.
(In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)
Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
$$
E: y^2 = x^3-36288x+2285712
$$
This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.
Sample solutions.
The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
$$
(-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
$$
Converting to $(u,v,w)$, we get
$$
(frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
$$
Normalizing, the first three solutions are
$$
beginalign
(95,-42,-13),\
(-35917476, -6829645, -10182731)\
(54059017558123943, -5902844861231317, -35013190193908290)
endalign
$$
Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
$$
(35917476, 6829645, 10182731)
$$
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.
Some solutions for small $t$'s.
beginarrayc
hline
t& u & v & w \ hline
2 & 35917476 & 6829645 & 10182731\ hline
3 & 1137565 & 647349 & 30196 \ hline
4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
endarray
It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.
for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.
For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)
In particular at $t=6$ the original curve
$$
frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
$$
is birational to
$$
y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
$$
via
$$
beginalign
(u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
endalign
$$
We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.
Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.
$t=2$ example
For $t=2$, we want integer solutions to
$$
f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
$$
We may freely multiply a rational $tneq 0$ to get
$$
tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
$$
Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).
Let
$$
beginalign
t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
endalign
$$
then it may be verified that this transform the original equation to
$$
g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
$$
which is a projective elliptic curve.
(In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)
Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
$$
E: y^2 = x^3-36288x+2285712
$$
This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.
Sample solutions.
The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
$$
(-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
$$
Converting to $(u,v,w)$, we get
$$
(frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
$$
Normalizing, the first three solutions are
$$
beginalign
(95,-42,-13),\
(-35917476, -6829645, -10182731)\
(54059017558123943, -5902844861231317, -35013190193908290)
endalign
$$
Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
$$
(35917476, 6829645, 10182731)
$$
Edit 1: Just noticed that the solution asks for $u,v,winmathbb N$, so while the search procedure is the same we require an extra check that they are $geq 1$.
Some solutions for small $t$'s.
beginarrayc
hline
t& u & v & w \ hline
2 & 35917476 & 6829645 & 10182731\ hline
3 & 1137565 & 647349 & 30196 \ hline
4 & 44334184670964 & 613899299195 & 18359866789309 \ hline
endarray
It appears that the torsion points correspond to a solution where one of $u,v,w$ is $0$, so we do not use them. Assuming this, solution to the original problem would imply finding points of infinite orders for an associated Elliptic curve.
for $t=5$ the curve is birational to an Elliptic curve of rank $0$ so there are no non-zero solutions. Hence not all $t$ yields a solution.
For $tgeq 6$ the curve parameters are huge enough that both Magma and Sage is unable to compute the generators. (Perhaps finding a solution would imply a better algorithm for finding generators?)
In particular at $t=6$ the original curve
$$
frac46656114791255(u^3 - v^3 - w^3 - 3^6 cdot 2uvw) = 0
$$
is birational to
$$
y^2z = x^3 - 1506290966208 x z^2 + 711559628544550032 z^3
$$
via
$$
beginalign
(u,v,w) &= (frac-1458 x+y+1033121268 z216, frac1458 x+y-1033121268 z216 , frac-x-2125764 z36)
endalign
$$
We can systematically try to solve for each $t>1$ and it seems like sometimes the equation will be birational to a rank 1 elliptic curve, would would mean infinitely many points.
Edit 1: Torsion points seems to corresponds to solutions where $u,v$ or $w=0$. On the other hand, for each $1leq tleq 20$, the resulting Elliptic curve each have two torsion points, would in turn means two solutions in the original equation. It looks like this may hold for each $t$, so at least two (coprime) solutions for each $t$.
$t=2$ example
For $t=2$, we want integer solutions to
$$
f(u,v,w) := u^3 - v^3 - w^3 - 18uvw=0
$$
We may freely multiply a rational $tneq 0$ to get
$$
tcdot f(u,v,w) = t(u^3 - v^3 - w^3 - 18uvw)=0
$$
Since scaling of $(u,v,w)$ to $(ku,kv,kw)$ also leads to a solution, it suffices to find any rational solution (then scaling it to integers if necessary).
Let
$$
beginalign
t&=-frac2^6cdot 3^65cdot 43=-frac46656215\
u&= frac-18x+y+1908z216, & v&= frac18 x + y - 1908 z216, & w &= frac-x-324216,
endalign
$$
then it may be verified that this transform the original equation to
$$
g(x,y,z):=x^3 - y^2 z - 36288 x z^2 + 2285712 z^3=0
$$
which is a projective elliptic curve.
(In practice this was found using Sagemath's EllipticCurve_from_cubic() function.)
Due to the scaling, it suffices to set $z=1$ and find rational solutions $(x,y,1)$ instead, i.e. solving
$$
E: y^2 = x^3-36288x+2285712
$$
This is a rank 1 Elliptic curve, so there are infinitely many rational points $(x,y,1)$. Each of these leads to a rational solution for $f(u,v,w)=0$, so we know that there are infinitely many solutions.
Sample solutions.
The curve $E$ has rank 1 with generator $G=(-168,1908)$, so that it generates points
$$
(-168,1908), (frac13954082809 , -frac1538896356148877), (frac7075793089672624225 , frac2407808634844631317444509275375)cdots
$$
Converting to $(u,v,w)$, we get
$$
(frac953, -14, -frac133)),(-frac11972492148877,-frac6829645446631,-frac1921278427),(frac540590175581239433768014003481000,-frac59028448612313173768014003481000,-frac22500604198924214472100)cdots
$$
Normalizing, the first three solutions are
$$
beginalign
(95,-42,-13),\
(-35917476, -6829645, -10182731)\
(54059017558123943, -5902844861231317, -35013190193908290)
endalign
$$
Edit 1: We require solutions in $mathbb N$, so we look for positive triplets. Since we can multiply by $-1$, negative triplets work as well. Here we find the solution
$$
(35917476, 6829645, 10182731)
$$
edited Aug 31 at 7:53
answered Aug 30 at 7:53
Yong Hao Ng
2,7791219
2,7791219
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This is just an observation. If, for any $t in mathbb N$ we have a solution $x,y,z$ then $2^kx, 2^ky, 2^kz$ is also a solution for any $kinmathbb N$. Indeed, $(2^kx)^3-(2^ky)^3-(2^kz)^3=3^t cdot 2 cdot2^kx cdot 2^ky cdot 2^kzimplies x^3-y^3-z^3=3^t cdot 2xyz$
â Right
Aug 30 at 6:29
@Right: since eq. is homogeneous, so if, for any $tin mathbbN$ we have a solution $x,y,z$ then $ax,ay,az$ is also a solution for any $ain mathbbN$.
â Oleg567
Aug 30 at 6:34
@Oleg567 That is right, did not see it immediately.
â Right
Aug 30 at 6:36